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Suppose customer arrivals at a post office are modeled by a Poisson process N with intensity λ > 0. Let T1 be the time of the first arrival. Let t > 0. Suppose we learn that by time t there has been precisely one arrival, in other words, Nt = 1. What is the distribution of T1 under this new information? In other words, find the conditional probability P(T1 ≤ s|Nt = 1) for all s ≥ 0.

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Answer:

Step-by-step explanation:

We need to find the conditional probability P( T1 < s|N(t)=1 )  for all s ≥ 0

P( time of the first person's arrival < s till time t exactly 1 person has arrived )

= P( time of the first person's arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )

{ As till time t, we know that exactly 1 person has arrived, thus relevant values of s : 0 < s < t }

P( time of the first person arrival < s, till time t exactly 1 person has arrived ) / P(exactly 1 person has arrived till time t )

= P( exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )

P(exactly x person has arrived till time t ) ~ Poisson(kt) where k = lambda

Therefore,

P(exactly 1 person has arrived till time s )/ P(exactly 1 person has arrived till time t )

= [ kse-ks/1! ] / [ kte-kt/1! ]

= (s/t)e-k(s-t)

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